What is the center x^2+y^2-x-6y+9=0 ?

The center x^2+y^2-x-6y+9=0 is (1/2 ,3)

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center x^2+y^2-x-6y+9=0

center

x^{2} + y^{2} - x - 6 y + 9 = 0

(\frac{1}{2}, 3)

Solution steps

x^{2} + y^{2} - x - 6 y + 9 = 0

Rewrite

x^{2} + y^{2} - x - 6 y + 9 = 0

in the form of the standard circle equation

(x - \frac{1}{2})^{2} + (y - 3)^{2} = (\frac{1}{2})^{2}

Therefore the circle properties are:

(a, b) = (\frac{1}{2}, 3), r = \frac{1}{2}

And the center is:

(\frac{1}{2}, 3)

(x + 4)^{2} + (y + 3)^{2} = 17

(x - 1)^{2} + (y + 4)^{2} = 16

4 x^{2} + 16 x + 15 + 4 y^{2} + 6 y = 0

(x + 4)^{2} + (y + 3)^{2} = 25

(x - 1)^{2} + (y - 3)^{2} = 9